Four Capacitors on 500 Volts

solved Capacitance

Q. A network of four 10 μF $C_{o} = 10$ μF capacitors is connected to a 500 V $V_{n} = 500$ V supply: three of them in series $C_{r} = \frac{1}{\frac{3}{C_{o}}}$ , and the fourth across that row $C_{n} = C_{r} + C_{o}$ . What is the equivalent capacitance of the network, in μF?

Inputs

each

C_{o}

10 μF

supply

V_{n}

500 V

Outputs

equivalent

C_{n}

How it's solved

Solved with Capacitance

Given

$C_{o} = 10$ μF
$V_{n} = 500$ V
$C_{r} = \frac{1}{\frac{3}{C_{o}}}$ from “three of them in series”
$C_{n} = C_{r} + C_{o}$ from “the fourth across that row”

Find $C_{n}$ .

Working

$C_{o} = 10$ μF $= 1 \times 10^{−5}$ F
three of them in series $C_{r} = \frac{1}{\frac{3}{C_{o}}}$
$C_{r} = \frac{1}{\frac{3}{1 \times 10^{−5}}}$
$C_{r} = 3.33333333 \times 10^{−6}$ F
the fourth across that row $C_{n} = C_{r} + C_{o}$
$C_{n} = 3.33333333 \times 10^{−6} + 1 \times 10^{−5}$
$C_{n} = 13.33$ μF
$C_{n} = 1.33333333 \times 10^{−5}$ F $= 13.3333333$ μF
$C_{n} \approx 13.3$ μFto 3 significant figures

Conclusion

In series the capacitances add as reciprocals, so the row of three holds a third as much; across it, the fourth simply adds.

The web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb

1require_relative "../physics"23question :four_capacitors_on_500_volts do4  kind :numerical5  describes "A network of four %{each} capacitors is connected to a %{supply} supply: three of them in series, and " \6            "the fourth across that row. %{equivalent}, in μF?"7  given each: "10 μF"8  given supply: "500 V"9  asks equivalent: "What is the equivalent capacitance of the network"10end

solution.rb

1require_relative "../questions/four_capacitors_on_500_volts"2require_relative "../science/electrostatic_potential_and_capacitance/capacitance"34solve :four_capacitors_on_500_volts, using: Capacitance do5  objects :one, :row, :network67  take each: one.capacitance8  take supply: network.voltage910  deduce_from "three of them in series", row.capacitance => 1 / (3 / one.capacitance)11  deduce_from "the fourth across that row", network.capacitance => row.capacitance + one.capacitance1213  equivalent = network.capacitance14  answer equivalent:, in: "μF", figures: 31516  conclusion "In series the capacitances add as reciprocals, so the row of three holds a third as much; " \17             "across it, the fourth simply adds."18end

playground.rb

1require_relative "../solutions/four_capacitors_on_500_volts"23play_with :four_capacitors_on_500_volts do4  title "Four Capacitors on 500 Volts"5end

The laws it drew on

quantity	written	units	within	from
permittivity of free space	$ε_{0}$	F/m	—	Capacitance
capacitance	$C$	F	0 … ∞	Capacitance
charge	$Q$	C	—	Capacitance
voltage	$V$	V	—	Capacitance
energy stored	$U$	J	—	Capacitance
area	$A$	m²	—	Capacitance
plate separation	$d$	m	—	Capacitance
dielectric constant	$K$	—	—	Capacitance
charge of ground	$Q_{g}$	C	—
capacitance of ground	$C_{g}$	F	0 … ∞
voltage of ground	$V_{g}$	V	—
charge of one	$Q_{o}$	C	—
capacitance of one	$C_{o}$	F	0 … ∞
voltage of one	$V_{o}$	V	—
charge of row	$Q_{r}$	C	—
capacitance of row	$C_{r}$	F	0 … ∞
voltage of row	$V_{r}$	V	—
charge of network	$Q_{n}$	C	—
capacitance of network	$C_{n}$	F	0 … ∞
voltage of network	$V_{n}$	V	—
dielectric constant of ground	$K_{g}$	—	—
area of ground	$A_{g}$	m²	—
plate separation of ground	$d_{g}$	m	—
dielectric constant of one	$K_{o}$	—	—
area of one	$A_{o}$	m²	—
plate separation of one	$d_{o}$	m	—
dielectric constant of row	$K_{r}$	—	—
area of row	$A_{r}$	m²	—
plate separation of row	$d_{r}$	m	—
dielectric constant of network	$K_{n}$	—	—
area of network	$A_{n}$	m²	—
plate separation of network	$d_{n}$	m	—
energy stored of ground	$U_{g}$	J	—
energy stored of one	$U_{o}$	J	—
energy stored of row	$U_{r}$	J	—
energy stored of network	$U_{n}$	J	—

Q_{c} = C_{c} \cdot V_{c}

charge on a capacitor

C_{c} = \frac{K_{c} \cdot ε_{0} \cdot A_{c}}{d_{c}}

a parallel plate capacitor

U_{c} = \frac{C_{c} \cdot {V_{c}}^{2}}{2}

energy in a capacitor

Q_{g} = C_{g} \cdot V_{g}

charge on a capacitor(ground)

Q_{o} = C_{o} \cdot V_{o}

charge on a capacitor(one)

Q_{r} = C_{r} \cdot V_{r}

charge on a capacitor(row)

Q_{n} = C_{n} \cdot V_{n}

charge on a capacitor(network)

C_{g} = \frac{K_{g} \cdot ε_{0} \cdot A_{g}}{d_{g}}

a parallel plate capacitor(ground)

C_{o} = \frac{K_{o} \cdot ε_{0} \cdot A_{o}}{d_{o}}

a parallel plate capacitor(one)

C_{r} = \frac{K_{r} \cdot ε_{0} \cdot A_{r}}{d_{r}}

a parallel plate capacitor(row)

C_{n} = \frac{K_{n} \cdot ε_{0} \cdot A_{n}}{d_{n}}

a parallel plate capacitor(network)

U_{g} = \frac{C_{g} \cdot {V_{g}}^{2}}{2}

energy in a capacitor(ground)

U_{o} = \frac{C_{o} \cdot {V_{o}}^{2}}{2}

energy in a capacitor(one)

U_{r} = \frac{C_{r} \cdot {V_{r}}^{2}}{2}

energy in a capacitor(row)

U_{n} = \frac{C_{n} \cdot {V_{n}}^{2}}{2}

energy in a capacitor(network)

C_{r} = \frac{1}{\frac{3}{C_{o}}}

three of them in series

C_{n} = C_{r} + C_{o}

the fourth across that row